1.

If `x=log_(2a) a,y=log_(3a) 2a ` and `z=log_(4a) 3a` then prove that `xyz+1=2yz`

Answer» `1+xyz=1(log_(2a)a)(log_(3a)2a)(log_(4a)3a)`
`=1+(loga)/(log2a)(log2a)/(log3a)(log3a)/(log4a)`
`=1+(loga)/(log4a)`
`=log_(4a)4a+log_(4a)a`
`=log_(4a)4a^(2)=2log_(4a)2a`
`=2(log_(3a)2a)(log_(4a)3a)=2yz`


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